Intersection of circle and parabola and quadratic discriminant

Question

Circle $x^2+y^2=9$ and parabola $y^2=8x$

They intersect at two points and at the same x-coordinate. On solving we get a quadratic equation in x as $x^2+8x-9=0$. It should give one value of x, hence the discriminant should be zero like in the below question where it is zero but not here. Why?

But on another question where circle $(x-\frac{3}{2})^2+y^2=\frac{9}{4}$ and parabola $y^2=-(x-4)$, on solving their intersection gives quadratic equation $x^2-4x+4=0$. Here too the two intersection point has the same x coordinate and we get the discriminant as zero which is understandable unlike the first question where discriminant is not zero. Why?

Thanks.

Answer 1

We begin solving with simple substitution and using the quadratic equation to find $\,x.\quad$ We then discard complex solutions in favor of real ones. \begin{align*} x^2+y^2&=9,\, y^2=8x\\ \implies x^2+8x&=9\\ \implies x&\in\{ -9,1\}\\ \implies y^2&\in\{-72,8\}\\ \implies y&\in\{\pm6i\sqrt{2},\pm2\sqrt{2}\} \end{align*}

$$\therefore (x,y)\in\{(1,-2\sqrt{2}),(1,2\sqrt{2})\}$$

For the second equation(s), try expanding and then multiplying both sides by the LCM to eliminate fractions. The math will be easier from there.

Answer 2

The perplexity created by the solutions to these pairs of equations is caused by the fact that they have "lives beyond $ \ \mathbb{R} " , $ which is something not generally discussed in algebra and analytic geometry courses. Students in early courses are generally told to "toss out" results that give negative squares of coordinate values.

The resolution of your question comes in considering that these "conic" equations describe objects in a four-dimensional space. This is hinted at when we solve your first set of equations for $ \ y \ \ , \ $ rather than $ \ x \ : $

$$ y^2 \ = \ 8x \ \ \rightarrow \ \ x \ = \ \frac{y^2}{8} \ \ \Rightarrow \ \ x^2 \ + \ y^2 \ \ = \ \ \frac{y^4}{64} \ + \ y^2 \ \ = \ \ 9 \ \ , \ $$ which gives us a quartic equation with the four solutions $$ y^2 \ \ = \ \ -32 \ \pm \ 32·\sqrt{1 \ + \ \frac{9}{16}} \ \ = \ \ 32·\left( -1 \ \pm \ \frac54 \right) \ \ = \ \ 8 \ \ , \ \ -72 $$ $$ y \ \ = \ \ \pm 2 \sqrt2 \ \ , \ \ \pm \sqrt{-72} \ \ = \ \ \pm i·6 \sqrt2 \ \ . $$

The first pair of values (which we keep) is associated with $ \ x \ = \ 1 \ \ $ and the second pair (which we "discard") is associated with $ \ x \ = \ -9 \ . $

But all of these solutions are correct in the larger scheme of things. If we write the coordinates using complex numbers as $ \ x \ = \ \alpha + i \beta \ $ and $ \ y \ = \ \gamma + i \delta \ \ $ (with $ \ \alpha \ , \ \beta \ , \ \gamma \ , \ \delta \ $ being real numbers), our "conic equations" become $$ (\alpha^2 \ - \ \beta^2) \ + \ 2 \alpha \beta · i \ + \ (\gamma^2 \ - \ \delta^2) \ + \ 2 \gamma \delta · i \ - \ 9 \ \ = \ \ 0 \ \ \ \text{(the "circle")} $$ and $$ (\gamma^2 \ - \ \delta^2) \ + \ 2 \gamma \delta · i \ - \ 8·(\alpha + i \beta) \ \ = \ \ 0 \ \ \ \text{(the "parabola")} \ \ . $$

The two-dimensional "slice" of this four-dimensional geometric object that we more often look at involves only the real parts of the coordinate variables, so we are taking $ \ \beta \ = \ \delta \ = \ 0 \ $ to produce the equations that you've solved, $ \ \alpha^2 + \gamma^2 - 9 \ = \ 0 \ \ $ and $ \ \gamma^2 - 8 \alpha \ = \ 0 \ \ . \ \ $ The result $ \ \alpha \ = \ 1 \ $ leads to $ \ \gamma^2 \ = \ 8 \ \ , \ $ which has the two square-roots $ \ \gamma \ = \ \pm \sqrt8 \ \ $ (really, complex square-roots with zero imaginary parts) and we are taught to "leave it at that", since $ \ \alpha \ = \ -9 \ \Rightarrow \ \gamma^2 \ = \ -72 \ $ will not give real values for $ \ \gamma \ \ . $

But if we instead look at the "slice" that includes the imaginary part of the $ \ y-$coordinate by taking $ \ \beta \ = \ \gamma \ = \ 0 \ \ , \ $ this portion of the "conic equations" is $ \ \alpha^2 - \delta^2 - 9 \ = \ 0 \ \ $ and $ \ -\delta^2 - 8 \alpha \ = \ 0 \ \ . \ \ $ Here. $ \ \alpha \ = \ -9 \ \Rightarrow \ \delta^2 \ = \ -8·(-9) \ = \ 72 \ \ $ will give us real values $ \ \delta \ = \ \pm 6 \sqrt2 \ \ , \ $ which correspond to $ \ y \ = \ \pm i·6 \sqrt2 \ \ . $ (These we ignore in our earlier studies.)

We can see these four solutions in the graphs below. The graph at left is the (usual) "reals-only" plot of the circle and parabola; the graph on the right shows the imaginary parts of the $ \ y-$coordinate, in which the parabola now "opens to the left" and the "circle" appears as a hyperbola.

For your first pair of conics, there are four distinct solutions because $ \ x \ $ can be positive or negative, so $ \ y \ $ has two real square-root values and two imaginary square-root values. WIth your second pair, this is avoided because the (real) circle is tangent to the $ \ y-$axis, so $ \ x \ $ can only be non-negative. Solving the pair of equations produces

$$ y^2 \ = \ 4 - x \ \ \rightarrow \ \ x \ = \ 4 - y^2 $$ $$ \Rightarrow \ \ \left(x \ - \ \frac32 \right)^2 \ + \ y^2 \ \ = \ \ \frac94 $$ $$ \rightarrow \ \ x^2 \ - \ 3x \ + \ y^2 \ \ = \ \ (4 - y^2)^2 \ - \ 3·(4 - y^2) \ + \ y^2 \ \ = \ \ y^4 \ - \ 4y^2 \ + \ 4 \ \ = \ \ (y^2 - 2)^2 \ \ = \ \ 0 \ \ . $$ So we again have a quartic equation, but the solutions are two real square-roots $ \ y^2 \ = \ 2 \ \Rightarrow \ y \ = \ \pm \sqrt2 \ \ , \ $ each with multiplicity $ \ 2 \ \ . \ $

We won't run through the "complexification" of the coordinates in detail again. The real part of the conic equations is $ \ \gamma^2 + \alpha - 4 \ = \ 0 \ \ , \ \ \alpha^2 - 3 \alpha + \gamma^2 \ = \ 0 \ \ , \ $ which are the equations you solved for the two points $ \ (2 \ , \ \pm \sqrt2) \ \ . \ $ The equations which give the imaginary part of $ \ y \ $ are $ \ -\delta^2 + \alpha - 4 \ = \ 0 \ \ , \ \ \alpha^2 - 3 \alpha - \delta^2 \ = \ 0 \ \ ; \ $ these have no real solutions for $ \ \delta \ \ $ and so there are no imaginary parts for $ \ y \ \ . $

We see this also in the graphs below. The real part of the "parabola" has four intersections with the "circle" (two intersection points of multiplicity $ \ 2 \ ) \ , \ $ but the "imaginary parabola" with its "reversed direction of opening" does not intersect the "imaginary hyperbolic" portion of the "circle".